Challenging Integral

**simplependulum** · June 21st 2009, 11:55 PM

solve $\int \frac{dx}{ sin^5{x} + cos^5{x} }$

and

$\int \frac{dx}{ sin^7{x} + cos^7{x} }$

**NonCommAlg** · June 22nd 2009, 06:52 AM

Originally Posted by simplependulum

solve $\int \frac{dx}{ sin^5{x} + cos^5{x} }$ and $\int \frac{dx}{ sin^7{x} + cos^7{x} }$

i'll solve the first one only and i'll leave the second integral to you or whoever likes trouble! first we need the following magical identity:

$\frac{1}{\sin^5 x + \cos^5 x} = \frac{4}{5} \left[\frac{1}{\sin x + \cos x} + \frac{(\sin x + \cos x)((\sin x - \cos x)^2 - 2)}{(\sin x - \cos x)^4 - 4(\sin x - \cos x)^2 - 1} \right].$

now finding $\int \frac{dx}{\sin x + \cos x}$ is straightforward. to find $I= \int \frac{(\sin x + \cos x)((\sin x - \cos x)^2 - 2)}{(\sin x - \cos x)^4 - 4(\sin x - \cos x)^2 - 1} \ dx$ substitute: $\sin x - \cos x = t.$ then $(\sin x + \cos x) \ dx = dt$ and thus

$I=\int \frac{t^2-2}{t^4 - 4t^2 - 1} \ dt.$ finally let $a=2+\sqrt{5}$ and $b=2-\sqrt{5}$ to get: $I=\frac{1}{2} \left[\int \frac{dt}{t^2-a} + \int \frac{dt}{t^2 - b} \right].$ note that $b < 0.$ now everything is nice and easy and you're good to go!

Remark: i think this is true that for any $n \in \mathbb{N}$ we have $\frac{1}{\sin^{2n+1}x + \cos^{2n+1} x} = \frac{2n}{2n+1}\left[\frac{1}{\sin x + \cos x} + (\sin x + \cos x)R((\sin x - \cos x)^2) \right],$ where $R(z)$ is some rational function of $z$ .

actually i think $R(z)=\frac{p(z)}{q(z)},$ where both $p(z)$ and $q(z)$ are monic polynomials with integer coefficients! for example, in your problem: $p(z)=z-2, \ q(z)=z^2-4z-1.$ i also think that we

always have $\deg q(z) = 1+\deg p(z).$ these claims are either wrong or easy to prove!

... let's just leave it as a challenge for whoever has time and patience to try it.

by the way, the identity $\frac{1}{\sin^3x + \cos^3x}=\frac{2}{3} \left[\frac{1}{\sin x + \cos x} + \frac{\sin x + \cos x}{(\sin x - \cos x)^2 + 1} \right]$ shows that my claims are not based on nothing!

**simplependulum** · June 22nd 2009, 09:29 PM

My solution is to substitute $t = \sin{x}+\cos{x}$
so,
$dt = (\cos{x} - \sin{x} )dx$

$\sin{x} \cos{x} = \frac{t^2 - 1}{2}$

$dx = \frac{dx}{ \sqrt{ 2 - t^2 } }$

Although it looks very troublesome , it become quite well after separating it into fractions .

However , i think your method is better than what i am using .

Thread: Challenging Integral

LinkBack

Thread Tools

Display